Question: Solve for $x$ : $ 7|x + 6| + 3 = 3|x + 6| + 6 $
Answer: Subtract $ {3|x + 6|} $ from both sides: $ \begin{eqnarray} 7|x + 6| + 3 &=& 3|x + 6| + 6 \\ \\ { - 3|x + 6|} && { - 3|x + 6|} \\ \\ 4|x + 6| + 3 &=& 6 \end{eqnarray} $ Subtract ${3}$ from both sides: $ \begin{eqnarray} 4|x + 6| + 3 &=& 6 \\ \\ { - 3} &=& { - 3} \\ \\ 4|x + 6| &=& 3 \end{eqnarray} $ Divide both sides by ${4}$ $ \dfrac{4|x + 6|} {{4}} = \dfrac{3} {{4}} $ Simplify: $ |x + 6| = \dfrac{3}{4}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x + 6 = -\dfrac{3}{4} $ or $ x + 6 = \dfrac{3}{4} $ Solve for the solution where $x + 6$ is negative: $ x + 6 = -\dfrac{3}{4} $ Subtract ${6}$ from both sides: $ \begin{eqnarray} x + 6 &=& -\dfrac{3}{4} \\ \\ {- 6} && {- 6} \\ \\ x &=& -\dfrac{3}{4} - 6 \end{eqnarray} $ Change the ${ - 6}$ to an equivalent fraction with a denominator of $4$ $ x = - \dfrac{3}{4} {- \dfrac{24}{4}} $ $ x = -\dfrac{27}{4} $ Then calculate the solution where $x + 6$ is positive: $ x + 6 = \dfrac{3}{4} $ Subtract ${6}$ from both sides: $ \begin{eqnarray} x + 6 &=& \dfrac{3}{4} \\ \\ {- 6} && {- 6} \\ \\ x &=& \dfrac{3}{4} - 6 \end{eqnarray} $ Change the ${ - 6}$ to an equivalent fraction with a denominator of $4$ $ x = \dfrac{3}{4} {- \dfrac{24}{4}} $ $ x = -\dfrac{21}{4} $ Thus, the correct answer is $x = -\dfrac{27}{4} $ or $x = -\dfrac{21}{4} $.